Question
Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,2]
, a solution is:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]
本题难度Medium。
【复杂度】
时间 O(N) 空间 O(N)
【思路】
与[LeetCode]Subsets差不多,区别有两点:
- 有重复的数字
- 因为有重复,所以需要先进行排序
只要在循环中加入判断:if(i==index||nums[i-1]!=nums[i])
(19行)即可。
【代码】
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
//require
List<List<Integer>> ans=new LinkedList<>();
ans.add(new LinkedList<Integer>());
Arrays.sort(nums);
//invariant
helper(0,new LinkedList<Integer>(),nums,ans);
//ensure
return ans;
}
private void helper(int index,List<Integer> preList,int[] nums,List<List<Integer>> ans){
int size=nums.length;
//base case
if(index==size)
return;
for(int i=index;i<size;i++){
if(i==index||nums[i-1]!=nums[i]){
preList.add(nums[i]);
ans.add(new LinkedList<Integer>(preList));
helper(i+1,preList,nums,ans);
preList.remove(preList.size()-1);
}
}
}
}